- #1

Jameson

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I know that [tex]\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica.

Hint please. :yuck:

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- Thread starter Jameson
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- #1

Jameson

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I know that [tex]\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica.

Hint please. :yuck:

- #2

Hurkyl

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- #3

Jameson

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Hmmm...I'll think out loud here. If the integral has a square-root and is in the form of [itex]c^2-x^2[/itex] then x is one leg, c is the hypotenuse, and the other leg is the previously mentioned radical.

Is trig substitution with right triangles on the right track? Since the hypotenuse is [itex]\sqrt{a^2+b^2}[/itex], it seems that one leg might need to be [itex]\sqrt{\arcsin{(x)}}[/itex] Somehow I don't think I'm on track.

- #4

amcavoy

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Use integration by parts. Remember to let u=arcsin(x) and v=x.

u sub: InverseLogAlgebraicTrigExp

u sub: InverseLogAlgebraicTrigExp

- #5

Hurkyl

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Sheesh, just **give** him the answer, why don't ya? :grumpy:

- #6

Jameson

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apmcavoy said:Use integration by parts. Remember to let u=arcsin(x) and v=x.

u sub: InverseLogAlgebraicTrigExp

Ah, thank you! It's so simple.

And thank you Hurkyl as well. I still have lots to learn.

- #7

amcavoy

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I apologize Hurkyl

- #8

tongos

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- #9

- #10

professorlucky

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equate the arc sine to another variable e.g y.making it a sine fxn.e.g the arc sine of 0.5=30,while sine30=0.5.this will simplify the integral and further substitution will conclude it

- #11

Tide

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The sum of the integrals

[tex]\int \sin^{-1} x dx + \int \sin y dy[/tex]

is just the area of the bounding rectangle: [itex] x \times \sin^{-1} x[/itex]

Since [tex]\int \sin y dy = -\cos y + C[/tex] and [tex]\cos y = \cos \sin^{-1} x = \sqrt {1-x^2}[/tex] it follows that

[tex]\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C[/tex]

- #12

FlashStorm

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Hey, I am really sorry to arise dead threads from the past (which i have seen though google) but somehting really weird happened me when I tried to use integration by parts on arcsinx.

let me show you:

S(arcsinx)= {v'(x)=1} {u(x)=arcsinx}

xarcsinx-S(x*d(arcsinx))=

xarcsinx-S(x/(1-x^2)^0.5= {u(x)=x v'(x)=arcsinx}

xarcsinx-xarcsinx+S(arcsin)dx

==> S(arcsinx)=S(arcsinx)

:\

I know I have done something really stupid here, but please be easy on me since I started studying Integrals only three days ago.

In the second time I used integration by parts, do I miss something , is there another efficient choice of v and u?

Thanks in advance,

Aviv

p.s: I will edit it better to use normal math signs once I figure out how.

- #13

Gib Z

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For the second time you integrate by parts, swap your choices.

- #14

FlashStorm

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xarcsine x -(x^2/2)(1/(Sqrt(1-x^2)))-(x/4)(sqrt(1/(1-x^2))+1/4(arcsinx)

this isn't going anywhere :(

- #15

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I know that [tex]\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica.

Hint please. :yuck:

Make [itex] x=\sin t [/itex]. Then apply part integration on the resulting integral. It's just a way to avoid the simple solution of part integrating directly.

- #16

Office_Shredder

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After the first integration by parts, I would use a substitution

- #17

FlashStorm

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did it without subtition, only using integration by parts.

if you are interested what I did then you are welcome to tell me to write my solution.

Thanks guys :)

gg

- #18

josiahsavino

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u=arcsinx

du=dx/(1-x^2)^1/2

dv=dx

v=x

uv-ingegral vdu = xarsinx-integral x/(1-x^2)^1/2

use u substitution with u = 1-x^2 so du = -2x

than you get

xarsinx + (1-x^2)^1/2 + c

- #19

Liqorice

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g'(x)=1 g(x)=x

// S means integral

S arcsin x dx= x arcsin x - S xdx/radical(1-xsquare) = x arcsin x + S (radical(1-xsquare))'dx=

=x arcsin x + radical(1-xsquare) +C

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